Physics 12 Assignment KEY
Equilibrium & 2-D Dynamics
1. Define the following terms:
• Newton’s laws of motion - three fundamental laws of motion which are the basis of
Newtonian mechanics are: 1) an object will remain at rest or in straight-line motion unless
acted on by an outside force; 2) the acceleration of an object is proportional to the force acting
on it and inversely proportional to its mass; 3) for every action force on an object, the object
exerts and equal and opposite reaction force
• Friction - a force that acts to slow the motion of an object or keep it at rest
• Tension - the magnitude of the force exerted on and by a cable, rope, or string
• Static equilibrium - the state of an object when the vector sum of all of the forces acting on it
is zero and the sum of all of the torques acting on it is zero
• Translational equilibrium – a state of equilibrium where the vector sum of all forces is zero
• Rotational equilibrium - a state of equilibrium where the vector sum of all torques is zero
• Torque – the tendency of a force to rotate an object about an axis, defined by the product of
the force on a object and the lever (moment) arm (NOTE: torque is not force)
• Lever arm - the perpendicular distance between the line along which the force is acting and
the pivot point of a rotation
• Center of mass - the point at which an object can be balanced
• Fulcrum (or pivot point) - the point around which rotational motion occurs
2. a) A box sits at rest on a rough 30o inclined plane. Draw free body diagram, showing all the forces
acting on the box. b) How does the diagram change if the box were sliding down the plane? c)
How does the diagram change if the box were sliding up the plane?
(a)
(b)
(c)
In (a) the friction is static and opposes the impending motion down the plane.
In (b) the friction is kinetic and opposes the motion down the plane.
In (c) the friction is kinetic and opposes the motion up the plane.
3. A bungee jumper momentarily comes to rest at the bottom of the dive before she springs back
upward. At that moment is the bungee jumper in equilibrium? Explain.
The bungee jumper is not in equilibrium, because the net force on the jumper is not zero. If
the jumper were at rest and the net force were zero, then the jumper would stay at rest by
Newton’s 1st law. The jumper has a net upward force when at the bottom of the dive, and
that is why the jumper is then pulled back upwards.
4. A ladder, leaning against a wall, makes a 60o angle with the ground. When it is more likely to slip:
when a person stands on the ladder near the top or near the bottom? Explain.
When the person stands near the top, the ladder is more
likely to slip. In the accompanying diagram to the right,
the force of the person pushing down on the ladder, Mg,
causes a clockwise torque about the contact point with
the ground, with lever arm dx. The only force causing a
counterclockwise torque about that same point is the
reaction force of the wall on the ladder, FW . While the
ladder is in equilibrium, FW will be the same magnitude as
the frictional force at the ground, FGx. Since FGx has a
maximum value, FW will have the same maximum value,
and so FW will have a maximum counterclockwise torque
that it can exert. As the person climbs the ladder, their
lever arm gets longer and so the torque due to their
weight gets larger. Eventually, if the torque caused by the person is larger than the
maximum torque caused by FW , the ladder will start to slip – it will not stay in equilibrium.
5. An earthen retaining wall is shown in the figure (a) below. The earth, particularly when wet, can
exert a significant force F on the wall. (a) What force produces the torque to keep the wall upright?
(b) Explain why the retaining wall in the figure (b) below would be much less likely to overturn.
(a) If we assume that the pivot point of rotation is the lower left corner of the wall in the
picture, then the gravity force acting through the CM provides the torque to keep the
wall upright. Note that the gravity force would have a relatively small lever arm (about
half the width of the wall) and so the sideways force would not have to be particularly
large to start to move the wall.
(b) With the horizontal extension, there are factors that make the wall less likely to
overturn:
• The mass of the second wall is larger, and so the torque caused by gravity
(helping to keep the wall upright) will be larger for the second wall.
• The center of gravity of the second wall is further to the right of the pivot point
and so gravity exerts a larger torque to counteract the torque due to F.
• The weight of the ground above the new part of the wall provides a large
clockwise torque that helps to counteract the torque due to F.
6. A child slides down a slide with a 28° incline, and at the bottom her speed is precisely half what it
would have been if the slide had been frictionless. Calculate the coefficient of kinetic friction
between the slide and the child.
Examine both situations, one without friction and one with
friction:
Without friction:
Fnet = F = m • a = mg • sinθ
Without friction:
∴ a = g sinθ
With friction:
Fnet = F - F f = m • a = mg • sinθ
With friction:
∴ mg • sinθ - µk • mg • cosθ = m • a
a = g(sinθ − µk • cosθ )
“d” is the displacement of the child (i.e. Super Raccoon Mario)
down the slide, regardless of the presence of friction…
v f 2 = vi 2 + 2ad
∴d =
v f 2 − vi 2
v f 2 − 02
vf 2
=
=
2a
2a
2a
Since "d" is the same in both situations,
vf 2
set ratio
2a
solve for µk .
v friction 2
vnone
2
v friction
in each situation equal to each other and
vnone 2
=
2anone
2a friction
v friction 2
Note: the child in this
FBD is actually Racoon
Mario!
=
a friction
anone
2
(2v friction )
2
=
a friction
anone
1 a friction g(sinθ − µk • cosθ )
=
=
4
anone
g sin θ
∴ µ k = 0.399
7. In the design of a supermarket, there are to be several ramps connecting different parts of the
store. Customers will have to push grocery carts up the ramps, and it is obviously desired that this
not be too difficult. An engineer has done a survey and found that almost no complains if the force
required is no more than 50 N. Will a slope of 5o be too steep, assuming a 30-kg grocery carts (full
of groceries)? Assume friction (wheels against ground, wheel on the axles, and so on), can be
accounted for by a co-efficient µk = 0.10.
It must be determined if the sum of the forces down the ramp
are less than 50 N while moving at constant speed. If so, no
one should complain about moving their shopping cart up
the ramp.
Fnet = 50 N - F - F f = 0
50 N = F + F f = mg sin θ + µk mg cos θ
50 N = (30 kg)(9.81 m/s2 )(sin5o + 0.10 • cos5o ) = 54.96 N
50 N ≠ 54.96 N
Since the opposing forces are greater than 50 N, the ramp is
deemed too steep.
8. A 28.0 kg block is connected to an empty 1.00 kg bucket by a cord
running over a frictionless pulley (See figure to the right). The
coefficient of static friction between the table and the block is 0.450,
and the coefficient of kinetic friction between the table and the block is
0.320. Sand is gradually added to the bucket until the system just
begins to move. (a) Calculate the mass of sand added to the bucket.
(b) Calculate the acceleration of the system.
(a) The mass of the sand can be determined by examining the
forces at the moment the system is about to accelerate (a = 0).
For the sand bucket...
Fnet = FG1 - FT = 0
FG1 = FT = ( mbucket + msand ) • g
For the block...
Fnet = FT - F f = 0
FT = F f
For the Atwood's Machine, FT is the same throughout the connector,
thus...
F f = ( mbucket + msand ) • g
µ s mblock g = ( mbucket + msand ) • g
∴ msand = µ s mblock − mbucket = 11.6 kg
(b) The acceleration of the system is found by…
For the sand bucket...
Fnet = FG1 - F f = msystem • a
( mbucket + msand ) • g − µk mblock g = ( mbucket + msand + mbucket ) • a
∴a =
( mbucket + msand ) • g − µk mblock g
= 0.88 m/s2
( mbucket + msand + mbucket )
9. Find the center of mass of the three-mass system shown below. Assume the spheres are point
masses.
Choose the reference point, x = 0 m, at the leftmost particle. Thus,
10. The CM of an empty 1050-kg car is 2.50 m behind the front of the car. How far from the front of
the car will the CM be when two people sit in the front seat 2.80 m from the front of the car, and
three people sit in the back seat 3.90 m from the front? Assume that each person has a mass of
70.0 kg.
11. A square uniform raft, 18 m by 18 m, of mass 6800 kg, is used as a ferryboat. If three cars, each
of mass 1200 kg, occupy its NE, SE, and SW corners, determine the CM of the loaded ferryboat.
12. Three forces are applied to a tree sapling, as shown in the
figure below, to stabilize it. If F1 = 282 N and F2 = 355 N, find F3
in magnitude and direction.
From the force diagram for the sapling, we can write:
ΣFx = 0
∴ F1 − F2 sin 20 o − F3 cos α = 0
282 N - (355 N)(sin 20 o ) - F3 cosα = 0
F3 cosα = 161 N
ΣFy = 0
∴ F2 cos 20 o − F3 sin α = 0
F3 sinα = 334 N
Thus, we have:
F3 = (161 N) 2 + (334 N) 2 \
F3 = 370 N
 334 N 
o
 = 64
 161 N 
∴θ = 180 o - 64 o = 116 o
α = tan -1 
13. Calculate the torque about the front support of a diving board, in the
figure to the right, exerted by a 60-kg person 3.0 m from that support.
We choose the fulcrum at the front support of the diving board,
with positive torques clockwise. The person’s weight produces
a positive torque about the fulcrum:
τ = Fg ⋅ r = mg ⋅ r = (60 kg)(9.81 N/kg)(3.0 m) = 1800 N ⋅ m
14. Two cords support a chandelier in the manner shown in the figure
to the right except that the upper wire makes an angle of 45° with
the ceiling. If the cords can sustain a force of 1300N without
breaking, what is the maximum chandelier weight that can be
supported?
There are actually three forces to consider: F1, F2,and the
weight, Fg, of the chandelier (as it develops tension in to
connecting vertical cord).
If we look at the point where the cords come together, we can
express the equilibrium of that point as follows: (The vector sum of F1, F2, and Fg - where
is the weight of the chandelier)
In the x direction:
F1x + F2x + Fgx = 0
-F1cos(45o)+ F2x + 0 = 0
So we know that F2x = F2 (It has no y component) = F1cos(45o)
In the y direction
F1y + F2y + Fgy = 0
F1sin(45o)+ 0 - Fg = 0
So we know that
F1 = (Fg)/sin(45o)
And
F2 = F1cos(45o) = ((Fg)/sin(45o))cos(45o) = Fg [since sin(45o) = cos(45o)]
So F1 is always going to be larger than F2 and F3 which are both equal to the weight, since
sin(45o) is less than one, and F1 = (Fg)/sin(45o).
Since the maximum tension is 1300 N, set F1 equal to 1300 N:
In the x direction:
F1x + F2x + F3x = 0
-(1300 N)cos(45o)+ F2x + 0 = 0
-919.24 N + F2x + 0 = 0
F2x = 919.24 N
In the y direction
F1y + F2y + F3y = 0
(1300 N)sin(45o)+ 0 - Fg = 0
919.24 - Fg = 0
Fg = 919.24 N = 920 N
15. A shop sign weighing 215 N is supported by a uniform 135-N beam
as shown in the figure below. Find the tension in the guy wire and
the horizontal and vertical forces exerted by the hinge on the
beam.
We choose the fulcrum at the point where the horizontal beam
makes contact with the wall, called the hinge (see FBD below):
ΣFx = 0
∴ FhingeH − FTx = 0
FhingeH = FTx = FT cos θ
ΣFy = 0
∴ F hingeV + FTy − mg − Mg = 0
F hingeV = mg + Mg − FTy
F hingeV = mg + Mg − FT sin θ
Στ = 0
∴ τ hingeH + τ T − τ mg + τ Mg = 0
Since the lever arm at the hinge is zero, the torque due the hinge force is zero. Thus,
Στ = 0
τ T − τ mg − τ Mg = 0
τ T = τ mg + τ Mg
(FT )(1.35 m)(sin 41o ) = (135 N)(0.85 m) + (215 N)(1.70 m)
(135 N)(0.85 m) + (215 N)(1.70 m)
(1.35 m)(sin 41o )
FT = 542 N
FT =
Therefore,
FhingeH = FT cos θ = (542 N)(cos 41o ) = 409 N
FhingeV = 135 N + 215 N − (542 N)(sin41o ) = -5.7 N
16. Consider a ladder in the figure below, with a painter climbing up. If the mass of the ladder is 12.0
kg, the mass of the painter is 60.0 kg, and the ladder begins to slip at its base when she is 70% of
the way up the length of the ladder, what is the coefficient of static friction between the ladder and
the floor? Assume the wall is frictionless.
Choose the pivot point to be the point on the ground where the
ladder makes contact. The ladder is in equilibrium, so the net
torque and net force must be zero. By stating that the ladder is
on the verge of slipping, the static frictional force at the ground,
FGx, is at its maximum value and so:
FGx = µFGy.
Given the height of the ladder is 4.0 m and the distance from the
wall is 3.0 m, the angle that the ladder makes with the ground is
θ = tan-1(4.0 m/3.0 m) = 53.13o. The length of the ladder is L.
ΣFx = 0
∴ FGx − FW = 0
FGx = FW
ΣFy = 0
∴ FGy − mg − Mg = 0
FGy = 706.32 N
Στ = 0
∴τ G + τ W − τ mg − τ Mg = 0
Since the lever arm at the pivot point is zero, the torque due the pivot point force, Fs, is
zero. Thus,
Στ = 0
∴τ W − τ mg − τ Mg = 0
τ W = τ mg + τ Mg
1
(FW )( L)(sin θ ) = (mg )(cos θ )( L) + ( Mg )(cos θ )(0.7 L)
2
1
(mg )(cos θ )( L) + ( Mg )(cos θ )(0.7 L)
2
FW =
( L)(sin θ )
1
(mg )(cos θ )( ) + ( Mg )(cos θ )(0.7)
2
FW =
sin θ
1
FW = (mg )(cot θ )( ) + ( Mg )(cot θ )(0.7)
2
FW = 353.16 N
∴µ =
FGx
= 0 .5
FGy